1. The sum of two numbers exceeds their product by one. Their difference is equal to their product less five. Find the two numbers.
$$\text {Let the number be x and y}$$
$$x + y = xy - 1$$
$$y – x = xy - 5$$
$$x + y + 1 = xy$$
$$y – x + 5 + = xy$$
$$x + y + 1 = y – x + 5$$
$$2x = 4$$
$$x = 2$$
$$y – 2 + 5 = 2y$$
$$y = 3$$
2. Find the value of \(x\)
$$
\begin{align}
2^{(x-3)} \times 8^{(x+2)} = 128\\
\end{align}
$$
$$
\begin{align}
& 2^x \div 2^3 \times 8^x \times 8^2 = 128\\\\
& \text {Let the number \(2^x\) be y}\\\\
& \frac y8 \times y^3 \times 64 = 128\\\\
& \frac {8y^4}{8} = \frac {128}{8}\\\\
& y^4 = 16\\\\
& y^4 = 2^4\\\\
& y = 2\\\\
& \text {Remember y = \(2^x\)}\\\\
& 2^x = 2\\\\
& 2^x = 2^1\\\\
& x = 1
\end{align}
$$
3. Evaluate:
$$
\begin{align}
\frac{-12 \div (-3) \times 4 - (-15)}{-5 \times 6 \div 2 + (-5)}
\end{align}
$$
$$
\begin{aligned}
& \text {Use BODMAS to evaluate} \\\\
& \frac{-12 \div (-3) \times 4 + 15)}{-5 \times 6 \div 2 - 5} \\\\
& \frac{4 \times 4 + 15}{-5 \times 3 - 5} \\\\
& \frac{16 + 15}{-15 - 5} \\\\
& \frac{31}{-20} \\\\
& -1\frac{11}{20}
\end{aligned}
$$
4. Without using a calculator/mathematical tables, evaluate leaving your answer as a simple fraction
$$
\begin{align}
\frac{(-4)(-2) + (-12) \div (+3)}{-9 - (15)} + \frac {-20 + (+4) + -6}{46 - (8+2) - 3}\\
\end{align}
$$
$$
\begin{aligned}
& \text {Use BODMAS to evaluate} \\\\
& \frac{8 -12 \div 3}{-24} + \frac {-20 + 4 - 6}{46 - 10 - 3}\\\\
& \frac{8 - 4}{-24} + \frac {-22}{33}\\\\
& \frac{4}{-24} + \frac {-2}{3}\\\\
& - \frac{1}{6} - \frac {2}{3}\\\\
& - \frac{5}{6}\\\\
\end{aligned}
$$
5. Evaluate
$$
\begin{align}
\frac{-8 \div 2 + 12 \times 9 - 4 \times 6}{56 \div 7 \times 2}\\\\
\end{align}
$$
$$
\begin{aligned}
& \text {Use BODMAS to evaluate} \\\\
& \frac{-4 + 108 - 24}{8 \times 2}\\\\
& \frac{80}{16}\\\\
& 5
\end{aligned}
$$
